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Q.
If $\sin^{-1} x + \sin^{-1} y +\sin^{-1}z = \pi$ then $ x \sqrt{1 -x^{2} } + y \sqrt{1-y^{2}} + z \sqrt{1-z^{2}} = $
Inverse Trigonometric Functions
Solution:
Let $\sin^{-1} x = A , \sin^{-1} y = B , \sin^{-1} z =C$ then A + B+ C = $\pi$ and then
$ \sin2A + \sin B + \sin2C = 4 \sin A \sin B \sin C$
4\
$ \Rightarrow 2x \sqrt{1-x^{2} } + 2y \sqrt{1-y^{2}} + 2 z \sqrt{1-z^{2}} $
$= 4xyz$
$ \Rightarrow x \sqrt{1-x^{2}} + y \sqrt{1-y^{2}} + z \sqrt{1-z^{2}} = 2xyz $